A firm is evaluating two machines. The first costs $250,000 and will require annual maintenance of $30,000 per year for 10 years. At the end of 10 years, the salvage value will be $75,000.

Question

The second machine costs $400,000 and will require $225,000 at the end of the fifth year. The salvage value after 10 years will be $175,000. Which machine should the firm select if interest in 8.5% compounded annually?

Answer 1

cost of 1st machine
= 250000 + 30000( 1 - 1.085^-10)/.085 - 75000(1.085)^-10
= 413,669.04

cost of 2nd machine
= 400000 + 225000(1.085)^-5 - 175000(1.085)^-10
= 472,235.27

1st is cheaper, I used the present time as my "focal point in time"
check my arithmetic